第1个回答 2020-01-09
∫(0->x+y) e^(-t^2) dt = ∫(0->x) x.(sint)^2 dt
x=0
∫(0->y) e^(-t^2) dt = 0
=> y=0
∫(0->x+y) e^(-t^2) dt = ∫(0->x) x.(sint)^2 dt
d/dx { ∫(0->x+y) e^(-t^2) dt } =d/dx { x. ∫(0->x) (sint)^2 dt }
(1 + dy/dx) . e^[ -(x+y)^2] = ∫(0->x) (sint)^2 dt + x(sinx)^2
dy/dx = { - e^[ -(x+y)^2] +x(sinx)^2 +∫(0->x) (sint)^2 dt } /e^[ -(x+y)^2]
dy/dx | x=0
=dy/dx | (x,y)=(0,0)
=( -1 +0 +0 ) /1
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